Question: Graph this system of equations and solve. $2x-4y = 12$ $y = -\dfrac{7}{2} x + 5$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Answer: Convert the first equation, $2x-4y = 12$ , to slope-intercept form. $y = \dfrac{1}{2} x - 3$ The y-intercept for the first equation is $-3$ , so the first line must pass through the point $(0, -3)$ The slope for the first equation is $\dfrac{1}{2}$ . Remember that the slope tells you rise over run. So in this case for every $1$ position you move up You must also move $2$ positions to the right. $2$ positions to the right. Graph the blue line so it passes through $(0, -3)$ and $(2, -2)$ The y-intercept for the second equation is $5$ , so the second line must pass through the point $(0, 5)$ The slope for the second equation is $-\dfrac{7}{2}$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move down (because it's negative) You must also move $2$ positions to the right. $2$ positions to the right. $7$ positions down from $(0, 5)$ is $(2, -2)$ Graph the green line so it passes through $(0, 5)$ and $(2, -2)$ The solution is the point where the two lines intersect. The lines intersect at $(2, -2)$.